Question: Find the range of $f(x) = \sin^4 x - \sin x \cos x  +\cos^4 x.$
Answer: We know that $\sin^2 x + \cos^2 x = 1.$  Squaring, we get
\[\sin^4 x + 2 \sin^2 x \cos^2 x + \cos^4 x = 1.\]Hence,
\begin{align*}
f(x) &= (\sin^4 x + \cos^4 x) - \sin x \cos x \\
&= (1 - 2 \sin^2 x \cos^2 x) - \sin x \cos x \\
&= 1 - \frac{1}{2} \sin 2x - \frac{1}{2} \sin^2 2x \\
&= \frac{9}{8} - \frac{1}{2} \left( \sin 2x + \frac{1}{2} \right)^2.
\end{align*}Since the range of $\sin x$ is $[-1,1],$ the range of $f(x)$ reaches a minimum when $\sin 2x = 1,$ in which case $f(x) = 0,$ and a maximum when $\sin 2x = -\frac{1}{2},$ in which case $f(x) = \frac{9}{8}.$  Therefore, the range of $f(x)$ is $\boxed{\left[ 0, \frac{9}{8} \right]}.$